Question

Complete the table to vertically stretch the parent function f(x)=x3
by a factor of 3.(2 points)
x y=kf(x)=
x3
0 0
1
2

Answer 1

To vertically stretch the parent function \( f(x) = x^3 \) by a factor of 3, we use the transformation formula \( y = kf(x) \), where \( k \) is the stretching factor. In this case, \( k = 3 \).

So, the new function will be:

\[
y = 3f(x) = 3x^3
\]

Now we can calculate the values for different \( x \) values. We already have \( x = 0 \):

- For \( x = 0 \):
\[
y = 3(0^3) = 3(0) = 0
\]

- For \( x = 1 \):
\[
y = 3(1^3) = 3(1) = 3
\]

- For \( x = 2 \):
\[
y = 3(2^3) = 3(8) = 24
\]

Now we can fill in the table:

\[
\begin{array}{|c|c|}
\hline
x & y = kf(x) = 3x^3 \\
\hline
0 & 0 \\
1 & 3 \\
2 & 24 \\
\hline
\end{array}
\]

So the completed table is:

\[
\begin{array}{|c|c|}
\hline
x & y = 3x^3 \\
\hline
0 & 0 \\
1 & 3 \\
2 & 24 \\
\hline
\end{array}
\]